FP3 - De Moivre

Hello!!

Please can anyone help me with this question? :

''Find tan3x in terms of tanx''. I know I'm meant to use De Moivre's theorem. I tried to do (cosx + isinx)^3 = isin3x + cos3x and then gather up the real and imaginary parts (cos3x = 4cos^3x - 3cosx, sin3x = 3sinx -4sin^3x). Then I put sin3x/cos3x and tried to mess about with the rest :P Unfortunately can't seem to get it all in terms of tanx :frown:

We really appreciate any help, thank you!
x

Reply 1

dvs

Here's a hint: don't simply your expressions for cos(3x) and sin(3x).

Of course it can still be done with your simplification if you notice that 1/cos^2(x) = sec^2(x) = 1 + tan^2(x) (after you divide top and bottom by cos^3(x)!).

Reply 2

anonpls123

dvs

Hey thanks!!
Was the ''of course..'' bit following on from the hint? Because if it wasn't, I still can't quite see why...sooorrrrrrrrryyyy gaah can't believe I want to do this at Uni and am getting stuck on this stuff already!!

Shall rept, take care
xx

Reply 3

dvs

Sorry about the confusion!

My second line doesn't follow from the hint. Basically what I was hinting at is that it would be better if you left the terms in the binomial expansion as they were, without simplifying and turning cos into sin and vice versa. It should be more apparent that you get tans if you do it this.

On the other hand, if you do decide to stick with your simplification, you can still get tans. I'll use s, c, t to denote sin(x), cos(x), tan(x), respectively.

tan(3x) = sin(3x)/cos(3x) = (3s - 4s^3)/(4c^3 - 3c)

Now if we divide top and bottom by c^3, we get
(3t/c^2 - 4t^3)/(4 - 3/c^2)

Note that 1/c^2 = sec^2(x) = 1 + tan^2(x) = 1 + t^2. So,
tan(3x) = (3t/c^2 - 4t^3)/(4 - 3/c^2)
= [3t(1+t^2) - 4t^3]/[4 - 3(1+t^2)]
etc.

Hopefully all is clear now.

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